What Is the Magnetic Field at Point P Due to a Short Current Element?

[reed2100]

Homework Statement

A short current element dl⃗ =(0.500mm)j^carries a current of 8.90A in the same direction asdl⃗ . Point P is located at r⃗ =(−0.730m)i^+(0.390m)k^.

Find the magnetic field at P produced by this current element.
Enter the x, y, and z components of the magnetic field separated by commas.2. Homework Equations

For a wire : dB = (μ naught/ 4π) (I) (dL X r→) / (r^2)

The Attempt at a Solution

dL = .0005 meters in the j→ direction

I = 8.9 amps in j→ direction

point P is located at r→ = -.730 meters i→ + .390 meters k→

so r is the magnitude of the distance between the point and the wire, so r = .73^2 + .39^2 = .685

μ naught / 4 pi = 10^-7 / 4 pi = 7.95 * 10^-9

Now I'm not sure what to do exactly. I know the cross product of dl and r→ is the determinant, and I know how to do that. Doing that I get a result of .000195 i + .000365 k.

Now do I take (7.95*10 ^-9) * (8.9 amps) (either .000195 or .000365) / (.685 meter^2) to find the i(x) and k(z) components of dB respectively? I've tried things like that and I couldn't get the right answer.

Note: I always do my calculations in standard units like meters and tesla first then convert at the end so I don't confuse myself so I don't think that was my mistake. I think I'm just not understanding what term actually means what in the equation, or how to work with vectors correctly. A nudge in the right direction would be appreciated, thanks for any and all advice.The correct answer is dBx, dBy, dBz = .306, 0 ,573 nT

 

[BvU]

Pretty good work!
I found one little flaw in the relevant equation: it's either $\displaystyle I\;d{\bf \vec l} \times {\bf \vec r}\over \displaystyle |r|^3$ or $\displaystyle I\; d{\bf \vec l} \times {\bf \hat r}\over \displaystyle |r|^2$

One little flaw in $\mu_0$: $\ \ \mu_0 = 4\pi \; 10^{-7}\ \Rightarrow \ \mu_0/(4\pi) = 10^{-7}$

And one corny little flaw in r = .73^2 + .39^2 = .685 See it ?

Fix them and you end up spot on !

 

[reed2100]

Thank you for responding!

So for the biot savart law I have a question. The first one you posted has dl→ × r→ whereas the second one uses r with a unit vector hat, so just to clarify, the first equation would be used if I had a vector r from my current element to the point of interest in terms of x, y, and z correct? And the 2nd one would be if I was given a vector in terms of a single unit vector? I'm confused on that part, I'm a little fuzzy on what I remember from vectors. Can you explain the difference between those two and why I would use one or the other?

The μ naught issue I see I made a mistake in the equation now; I must have misread that constant relationship at first.

But the last flaw is also confusing. I know the magnitude of a vector is the square root of the sum of the squared components, so I thought : r^2 = x(i) ^2 + y(j) ^2 , in the original post I accidentally said r instead of r^2. But I can see that it would be an error regardless if I originally chose the r^3 biot savart equation. So I don't see where I went wrong in that equation with what I chose originally, regarding the r^2.

 

[Feodalherren]

They are one and the same remember that "r-hat" is just R divided by it's own magnitude. So in one of the cases one magnitude has been factored out and canceled and in the other not. They are exactly the same.

\vec R = |R| Rhat

You would use the one with Rhat if you didn't know the distance but knew the direction (it might put you a step closer even though you can't get a numerical answer). If you know the just the distance (R) but not the direction there is no sense in calculating Rhat so then you would use the r^3 version.